以下哪一种技术是整数除2的最佳选择,为什么?

技巧1:

x = x >> 1;

技巧2:

x = x / 2;

这里x是整数。


当前回答

使用除法(/),假设这样更清楚。编译器将相应地进行优化。

其他回答

使用最能描述您要做的事情的操作。

如果你将数字作为一个比特序列来处理,请使用bitshift。 如果你把它当作一个数值,使用除法。

请注意,它们并不完全相等。对于负整数,它们可以给出不同的结果。例如:

-5 / 2  = -2
-5 >> 1 = -3

(ideone)

x / 2更清晰,x >> 1快不了多少(根据一个微基准测试,对于Java JVM快30%左右)。正如其他人所注意到的,对于负数,舍入略有不同,所以当您想处理负数时必须考虑这一点。一些编译器可能会自动将x / 2转换为x >> 1,如果他们知道这个数字不可能是负数(即使我无法验证这一点)。

即使x / 2可能不使用(慢)除法CPU指令,因为一些捷径是可能的,但它仍然比x >> 1慢。

(This is a C / C++ question, other programming languages have more operators. For Java there is also the unsigned right shift, x >>> 1, which is again different. It allows to correctly calculate the mean (average) value of two values, so that (a + b) >>> 1 will return the mean value even for very large values of a and b. This is required for example for binary search if the array indices can get very large. There was a bug in many versions of binary search, because they used (a + b) / 2 to calculate the average. This doesn't work correctly. The correct solution is to use (a + b) >>> 1 instead.)

显然,如果你是在为下一个阅读你的代码的人写代码,那么你应该追求“x/2”的清晰度。

然而,如果速度是你的目标,那就两种方法都试一试,把握好时间。几个月前,我做了一个位图卷积例程,它涉及到在一个整数数组中步进,并将每个元素除以2。我做了各种各样的事情来优化它,包括用“x>>1”代替“x/2”的老技巧。

当我计算这两种方式时,我惊奇地发现x/2比x>>1快

这是使用Microsoft VS2008 c++并打开默认优化。

X/Y是正确的…和" >> "移位运算符..如果我们想要二除一个整数,我们可以使用(/)被除数运算符。移位运算符用于移位位。

x = x / 2; x / = 2;我们可以这样用…

查看编译器的输出来帮助您做出决定。我在x86-64上使用 gcc (gcc) 4.2.1 20070719 [FreeBSD]

也可以在godbolt上看到编译器的在线输出。

What you see is the compiler does use a sarl (arithmetic right-shift) instruction in both cases, so it does recognize the similarity between the two expressions. If you use the divide, the compiler also needs to adjust for negative numbers. To do that it shifts the sign bit down to the lowest order bit, and adds that to the result. This fixes the off-by-one issue when shifting negative numbers, compared to what a divide would do. Since the divide case does 2 shifts, while the explicit shift case only does one, we can now explain some of the performance differences measured by other answers here.

C代码与汇编输出:

对于除法,你的输入是

int div2signed(int a) {
  return a / 2;
}

这个编译成

    movl    %edi, %eax
    shrl    $31, %eax            # (unsigned)x >> 31
    addl    %edi, %eax           # tmp = x + (x<0)
    sarl    %eax                 # (x + 0 or 1) >> 1  arithmetic right shift
    ret

shift也是一样

int shr2signed(int a) {
  return a >> 1;
}

输出:

    sarl    %edi
    movl    %edi, %eax
    ret

其他isa即使不能做得更好,也能同样有效地做到这一点。例如GCC For AArch64使用:

        add     w0, w0, w0, lsr 31      // x += (unsigned)x>>31
        asr     w0, w0, 1               // x >>= 1
        ret